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Can you give a simple example of a closed orientable manifold with an everywhere nonzero section of its tangent bundle, but where the tangent bundle is not trivial?

If the manifold is not orientable, it is not too hard to come up with examples (e.g. Klein bottle). But all the obvious constructions I can think of fail when you require orientability. For example, a natural thing to do is to take a product of a parallelizable manifold (say $S^1$) with a manifold with no everywhere nonzero section of its tangent bundle. But there is no guarantee that this will have nontrivial tangent bundle: for example, $S^1\times S^n$ is parallelizable for all $n$.

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  • $\begingroup$ Huh. I hadn't thought about it before, but that's interesting that $S^1 \times S^n$ is parallelizable. $\endgroup$
    – Simon Rose
    Dec 17 '14 at 13:28
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The sphere $S^5$ is closed, orientable, and there is an explicit nonvanishing section of its tangent bundle: $(x_2, -x_1, x_4, -x_3, x_6, -x_5)$ (if you view $S^5 \subset \mathbb{R}^6$ in the standard way). But Adams' theorem on the Hopf invariant says that the only parallelizable spheres are $S^0$, $S^1$, $S^3$, and $S^7$. (I chose $S^5$, but any odd sphere other than these four examples works.)

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    $\begingroup$ I cited Adams, but I think this is a corollary of Adams' theorem that was already known before. Wikipedia says this is due to Kervaire and Bott--Milnor. $\endgroup$ Dec 17 '14 at 13:28
  • $\begingroup$ Very good, thanks! It's interesting that there don't seem to be any examples below 5 dimensions. $\endgroup$ Dec 21 '14 at 22:57
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By the Poincare-Hopf theorem and its converse, if $X$ is a closed connected smooth orientable manifold, then $X$ admits a nonvanishing vector field iff the Euler characteristic of $X$ is zero. So it suffices to find an example of such a manifold whose Euler characteristic is zero but such that at least one of its Stiefel-Whitney or Pontryagin classes does not vanish.

Such an example does not exist in low dimensions (and note that the Euler characteristic condition automatically holds in all odd dimensions by Poincare duality):

  • If $\dim X = 1$, then $X$ must be $S^1$, which is parallelizable.
  • If $\dim X = 2$, then $X$ must be $S^1 \times S^1$, which is parallelizable.
  • If $\dim X = 3$, then any closed oriented smooth $3$-manifold is parallelizable.

So $\dim X \ge 4$. Unfortunately I'm having some trouble writing down a $4$-dimensional counterexample, and I'm not sure if one exists.

But here are a bunch of $5$-dimensional counterexamples: let $X = S^1 \times Y$ where $Y$ is a closed connected smooth orientable $4$-manifold with nonzero signature, for example $\mathbb{CP}^2$. Then $X$ is closed, connected, smooth, orientable, and has Euler characteristic $0$, but by the signature theorem $p_1(Y) \neq 0$ and so $p_1(X) \neq 0$, so $X$ is not stably parallelizable.

More generally you can take $X = S^1 \times Y$ where $Y$ is closed, connected, smooth, orientable, but not stably parallelizable, e.g. because at least one of its Stiefel-Whitney or Pontryagin classes does not vanish. This construction requires that $\dim Y \ge 4$.

Edit: An easier example is to take $X = \mathbb{RP}^5$. Like $S^5$, this is closed, connected, smooth, orientable, and has Euler characteristic $0$, hence has a nonvanishing vector field (which I think can be constructed from a corresponding nonvanishing vector field on $S^5$). But its Stiefel-Whitney classes don't all vanish; for example, $w_2(\mathbb{RP}^5) \neq 0$. So $X$ is not stably parallelizable. This doesn't require any big theorems.

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  • $\begingroup$ Thanks for this, wish I could accept two answers! It's interesting that in three dimensions parallelizability is automatic. $\endgroup$ Dec 21 '14 at 23:02

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